Math 251 - Alexiades
Review for Exam 2 on Chap.4
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Concepts
vector space, subspace
linear independence, span, basis, dimension
row space, column space, nullspace, rank, nullity of m×n matrix
linear transformation T: V → W
one-to-one, onto transformations
Precise definition of
vector space (10 axioms), subspace: a subset of a v.s. which is itself a v.s. with the same operations.
linearly independent set {v1,v2,...,vk} in V :
c1v1+c2v2+...+ckvk=0 implies c1,c2,...,ck are all 0.
basis of V : a subset of V which is linearly independent and spans V.
dimension of V : the number of vectors in any basis of V.
linear transformation T: V → W :
T(c1x1+c2x2) = c1T(x1)+c2T(x2) ∀ scalars ci, ∀ xi∈V.
one-to-one mapping T: V → W :
T(u)=T(v) implies u=v ∀u,v ∈ V.
For a linear transformation: being one-to-one amounts to: T(x)=0 implies x=0,
i.e. nullity(T)=0, i.e. rank[T] = n
onto mapping T: V → W :
for each y ∈ W, y = T(x) for some x ∈ V,
i.e. T(x) = y has a solution x ∈ V for each y ∈ W
For a linear transformation: being onto amounts to: [T]x=y solvable ∀ y ∈ ℝm,
i.e. cols of [T] span ℝm, i.e. rank[T] = m
Methods: Know How To
decide if a set/subset (with given operations) constitutes a vector space/subspace
decide if a set {v1,v2,...,vk}
is linearly independent
express a vector as a linear combination of {v1,v2,...,vk}
decide if a set {v1,v2,...,vk}
spans a space/subspace
find a basis (and dimension) for a space/subspace
find a basis for row space, column space, nullspace of m×n matrix
decide if a transformation T: V → W is linear or not
find standard matrix for a linear transformation T: ℝn → ℝm
decide if a linear transformation T: V → W is one-to-one, if it is onto
Basic / crucial Theorems to know (...among others...)
Dimension Thm: rank(A) + nullity(A) = n
What rank tells us, for an m×n matrix A:
rank(A) = r
⇐⇒
the column space of A is an r-dimensional subspace of ℝm
⇐⇒
r of the n columns are linearly independent in ℝm
⇐⇒
the row space of A is an r-dimensional subspace of ℝn
⇐⇒
r of the m rows are linearly independent in ℝn
⇐⇒
nullity(A) = n−r
⇐⇒
Ax=0 has n−r linearly independent solutions
rank(A) = n
⇐⇒
nullity(A) = 0
⇐⇒
Ax=0 has only the trivial solution
⇐⇒
the columns of A are linearly independent in ℝm
⇐⇒
Ax=b has at most one solution ∀ b ∈ ℝm
rank(A) = m
⇐⇒
the columns of A span ℝm
⇐⇒
Ax=b is consistent ∀ b ∈ ℝm
Fundamental Thm for square matrices:
For a square n×n matrix A, the following are equivalent:
1. A is invertible
2. A ∼ In (i.e. A is row equivalent to the Identity)
3. A is expressible as a product of elementary matrices
4. rank(A) = n
5. det(A) ≠ 0
6. the homogeneous system Ax = 0 has only the trivial solution
7. nullity(A) = 0
8. the nonhomogeneous system Ax = b is consistent ∀ b ∈ ℝn
9. the nonhomogeneous system Ax = b has unique solution ∀ b ∈ ℝn
10. λ = 0 is not an eigenvalue of A
11. the columns [and rows] of A are linearly independent vectors in ℝn
12. the columns [and rows] of A span ℝn
13. the columns [and rows] of A form a basis for ℝn
14. the linear operator TA: ℝn → ℝn, defined by TA(x)=Ax, is one-to-one
15. the linear operator TA: ℝn → ℝn, defined by TA(x)=Ax, is onto